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Difference between revisions of "1973 IMO Problems/Problem 3"

m
Line 4: Line 4:
 
<math>x^4 + ax^3 + bx^2 + ax + 1 = 0</math>
 
<math>x^4 + ax^3 + bx^2 + ax + 1 = 0</math>
 
has at least one real solution. For all such pairs <math>(a, b)</math>, find the minimum value of <math>a^2 + b^2</math>.
 
has at least one real solution. For all such pairs <math>(a, b)</math>, find the minimum value of <math>a^2 + b^2</math>.
 +
  
 
==Solution==
 
==Solution==
−
Substitute <math>z=x+1/x</math> to change the original equation into <math>z^2+az+b-2=0</math>. This equation has solutions <math>z=\frac{-a \pm \sqrt{a^2+8-4b}}{2}</math>. We also know that <math>|z|=|x+1/x| \geq 2</math>. So,
 
  
−
<cmath>\left | \frac{-a \pm \sqrt{a^2+8-4b}}{2} \right | \geq 2</cmath>
+
Substitute <math>z=x+1/x</math> to change the original equation into <math>z^2+az+b-2=0</math>. This equation has solutions <math>z=\frac{-a \pm \sqrt{a^2+8-4b}}{2}</math>. We also know that
 +
 
 +
<cmath>|z|=|x+1/x| \geq 2.\ \ \ \ \ \ \ \ (1)</cmath>
 +
 
 +
So,
  
−
<cmath>\frac{|a|+\sqrt{a^2+8-4b}}{2} \geq 2</cmath>
+
<cmath>\left | \frac{-a \pm \sqrt{a^2+8-4b}}{2} \right | \geq 2\ \ \ \ \ \ \ \ (2)</cmath>
 +
 
 +
<cmath>\frac{|a|+\sqrt{a^2+8-4b}}{2} \geq 2\ \ \ \ \ \ \ \ \ \ (3)</cmath>
  
 
<cmath>|a|+\sqrt{a^2+8-4b} \geq 4</cmath>
 
<cmath>|a|+\sqrt{a^2+8-4b} \geq 4</cmath>
Line 16: Line 22:
 
Rearranging and squaring both sides,
 
Rearranging and squaring both sides,
  
−
<cmath>a^2+8-4b \geq a^2-16|a|+16</cmath>
+
<cmath>a^2+8-4b \geq a^2-16|a|+16\ \ \ \ \ \ \ \ (4)</cmath>
  
−
<cmath>2|a|-b \geq 2</cmath>
+
<cmath>2|a|-b \geq 2\ \ \ \ \ \ \ \ \ \ \ \ (5)</cmath>
  
−
So, <math>a^2+b^2 \geq a^2+(2-2|a|)^2 = 5a^2-8|a|+4 = 5(|a|-\frac{4}{5})^2+\frac{4}{5}</math>.
+
So,
 +
 
 +
<cmath>a^2+b^2 \geq a^2+(2-2|a|)^2\ \ \ \ \ \ \ \ (6)</cmath>
 +
 
 +
<cmath>= 5a^2-8|a|+4 = 5 \left( |a|-\frac{4}{5} \right )^2+\frac{4}{5}</cmath>.
  
 
Therefore, the smallest possible value of <math>a^2+b^2</math> is <math>\frac{4}{5}</math>, when <math>a=\pm \frac{4}{5}</math> and <math>b=\frac{-2}{5}</math>.
 
Therefore, the smallest possible value of <math>a^2+b^2</math> is <math>\frac{4}{5}</math>, when <math>a=\pm \frac{4}{5}</math> and <math>b=\frac{-2}{5}</math>.
  
 
Borrowed from [http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln733.html]
 
Borrowed from [http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln733.html]
 +
 +
 +
==Remarks (added by pf02, June 2025)==
 +
 +
1. The solution above is incomplete and it has some serious errors.
 +
The result is correct, but the method for obtaining it is flawed.
 +
For the sake of reference to the steps in the solution I added labels,
 +
but I did not make any changes to the solution.  I will highlight the
 +
errors and the missing steps:
 +
 +
A. (1) is true when we know that <math>x</math> is real.  By hypothesis the
 +
equation has at least one real solution, so (1) is true for this
 +
solution.  A necessary condition for this is that
 +
<math>a^2 + 8 - 4b \ge 0</math>.  We have to use this in the proof, or if we
 +
obtain a result without having used it, we need to verify that the
 +
result satisfies this condition.  This is missing from the solution.
 +
 +
B. (2) is equivalent to (3) when <math>\pm</math> is interpreted as an
 +
<math>\mathbf{"or"}</math>.  (If the <math>\pm</math> would be an <math>\mathbf{"and"}</math>
 +
then (2) and (3) would not be equivalent.)  In our problem the
 +
<math>\pm</math> is an <math>\mathbf{"or"}</math> so the proof can proceed.  However,
 +
while easy, going from (2) to (3) is not obvious, and it is an
 +
important step in the proof, so it should be explained.
 +
 +
C. There is a small computational error in (4): we should have
 +
<math>8|a|</math>, not <math>16|a|</math>. This error is in writing, the computation
 +
proceeds as if it did not happen.
 +
 +
D. Going from (5) to (6) is a very serious error.  Replacing
 +
<math>b^2</math> by <math>(2 - 2|a|)^2</math> can be done only when certain conditions
 +
are satisfied (for example if we know that <math>b \le 0</math> and
 +
<math>2 - 2|a| \ge 0</math> or if we know that <math>b = 2 - 2|a|</math>).  So
 +
writing down (6) is completely unjustified.
 +
 +
2. I will give a different solution below.
 +
 +
 +
==Solution 2==
 +
 +
 +
 +
 +
 +
 +
 +
 +
 +
 +
  
 
== See Also == {{IMO box|year=1973|num-b=2|num-a=4}}
 
== See Also == {{IMO box|year=1973|num-b=2|num-a=4}}

Revision as of 00:56, 15 June 2025

Problem

Let $a$ and $b$ be real numbers for which the equation $x^4 + ax^3 + bx^2 + ax + 1 = 0$ has at least one real solution. For all such pairs $(a, b)$, find the minimum value of $a^2 + b^2$.


Solution

Substitute $z=x+1/x$ to change the original equation into $z^2+az+b-2=0$. This equation has solutions $z=\frac{-a \pm \sqrt{a^2+8-4b}}{2}$. We also know that

\[|z|=|x+1/x| \geq 2.\ \ \ \ \ \ \ \ (1)\]

So,

\[\left | \frac{-a \pm \sqrt{a^2+8-4b}}{2} \right | \geq 2\ \ \ \ \ \ \ \ (2)\]

\[\frac{|a|+\sqrt{a^2+8-4b}}{2} \geq 2\ \ \ \ \ \ \ \ \ \ (3)\]

\[|a|+\sqrt{a^2+8-4b} \geq 4\]

Rearranging and squaring both sides,

\[a^2+8-4b \geq a^2-16|a|+16\ \ \ \ \ \ \ \ (4)\]

\[2|a|-b \geq 2\ \ \ \ \ \ \ \ \ \ \ \ (5)\]

So,

\[a^2+b^2 \geq a^2+(2-2|a|)^2\ \ \ \ \ \ \ \ (6)\]

\[= 5a^2-8|a|+4 = 5 \left( |a|-\frac{4}{5} \right )^2+\frac{4}{5}\].

Therefore, the smallest possible value of $a^2+b^2$ is $\frac{4}{5}$, when $a=\pm \frac{4}{5}$ and $b=\frac{-2}{5}$.

Borrowed from [1]


Remarks (added by pf02, June 2025)

1. The solution above is incomplete and it has some serious errors. The result is correct, but the method for obtaining it is flawed. For the sake of reference to the steps in the solution I added labels, but I did not make any changes to the solution. I will highlight the errors and the missing steps:

A. (1) is true when we know that $x$ is real. By hypothesis the equation has at least one real solution, so (1) is true for this solution. A necessary condition for this is that $a^2 + 8 - 4b \ge 0$. We have to use this in the proof, or if we obtain a result without having used it, we need to verify that the result satisfies this condition. This is missing from the solution.

B. (2) is equivalent to (3) when $\pm$ is interpreted as an $\mathbf{"or"}$. (If the $\pm$ would be an $\mathbf{"and"}$ then (2) and (3) would not be equivalent.) In our problem the $\pm$ is an $\mathbf{"or"}$ so the proof can proceed. However, while easy, going from (2) to (3) is not obvious, and it is an important step in the proof, so it should be explained.

C. There is a small computational error in (4): we should have $8|a|$, not $16|a|$. This error is in writing, the computation proceeds as if it did not happen.

D. Going from (5) to (6) is a very serious error. Replacing $b^2$ by $(2 - 2|a|)^2$ can be done only when certain conditions are satisfied (for example if we know that $b \le 0$ and $2 - 2|a| \ge 0$ or if we know that $b = 2 - 2|a|$). So writing down (6) is completely unjustified.

2. I will give a different solution below.


Solution 2

See Also

1973 IMO (Problems) • Resources
Preceded by
Problem 2 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 4
All IMO Problems and Solutions
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